Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{6x + 60}{x - 6} \div \dfrac{5x^2 + 50x}{x^2 - 8x + 12} $
Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{6x + 60}{x - 6} \times \dfrac{x^2 - 8x + 12}{5x^2 + 50x} $ First factor the quadratic. $n = \dfrac{6x + 60}{x - 6} \times \dfrac{(x - 6)(x - 2)}{5x^2 + 50x} $ Then factor out any other terms. $n = \dfrac{6(x + 10)}{x - 6} \times \dfrac{(x - 6)(x - 2)}{5x(x + 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 6(x + 10) \times (x - 6)(x - 2) } { (x - 6) \times 5x(x + 10) } $ $n = \dfrac{ 6(x + 10)(x - 6)(x - 2)}{ 5x(x - 6)(x + 10)} $ Notice that $(x + 10)$ and $(x - 6)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 6\cancel{(x + 10)}(x - 6)(x - 2)}{ 5x\cancel{(x - 6)}(x + 10)} $ We are dividing by $x - 6$ , so $x - 6 \neq 0$ Therefore, $x \neq 6$ $n = \dfrac{ 6\cancel{(x + 10)}\cancel{(x - 6)}(x - 2)}{ 5x\cancel{(x - 6)}\cancel{(x + 10)}} $ We are dividing by $x + 10$ , so $x + 10 \neq 0$ Therefore, $x \neq -10$ $n = \dfrac{6(x - 2)}{5x} ; \space x \neq 6 ; \space x \neq -10 $